Remainder Theorem

A cubic function f(x) leaves a remainder of -5 and 7 when it is divided by x + 1 and x - 2 respectively.

Find the remainder when the cubic function f(x) is divided by x^2 - x - 2.

Thanks for the help.

A cubic function f(x) leaves a remainder of -5 and 7 when it is divided by x + 1 and x - 2 respectively.

Find the remainder when the cubic function f(x) is divided by x^2 - x - 2.

Thanks for the help.

Let the remainder be ax+b

Function = Quotient x Divisor + Remainder

f(x) = Q(x) [x^2-x-2) + (ax+b)

f(-1) = -5

-5 = Q(-1) [0] + (-a+b)

a-b =5

f(2) = 7

7 = Q(2) [0] + (2a+b)

2a +b = 7

3a = 12

a= 4

4-b = 5

b = -1

Remainder = 4x-1

Note: anyhow solve one... check for carelessness.

The fun part of this question is that the cubic function is not given and yet the

remainder can be found when the unknown cubic function is divided by x^2 - x - 2.

In addition, since x + 1 is a factor of x^2 - x - 2, when the function is divided by

x +1, the remainder can be found by substituting x = - 1 into the remainder 4x -1

,whereby the function is divided by x^2 - x - 2, ie 4(-1) - 1 = - 5

Similarly, since x - 2 is a factor of x^2 - x - 2, when the function is divided by

x - 2, the remainder can be found by substituting x = 2 into the remainder 4x -1

,whereby the function is divided by x^2 - x - 2, ie 4(2) - 1 = 7.